Homework Helpers: Physics

5 Electric Charges, Forces, and Fields

In Chapter 3 we discussed gravitational potential energy. We said that when a person lifts an object up, he or she is doing work against the gravitational field (or “gravity” for short) around Earth. At its new height, the object would have more gravitational potential energy, with reference to the floor. When an object falls, the gravitational field does work on the object. This work exerts a force over a distance, and the object”s gravitational potential energy is changed into kinetic energy as it falls. If you can transfer this entire mental model over to the electric charges and force, you will be on your way to understanding electric potential energy.

It is really quite simple. Electric potential energy is the energy stored by an electric charge due to its position in an electric field. Imagine holding a positive test charge near a stationary negatively charged object. The electric field is directed toward the negatively charged object, indicating the direction of force that a positively charged object would experience within it. The positive test charge that we hold has a certain amount of electric potential energy at this point. If we release it, the field around the negative charge does work on our positive test charge, and attracts it. The electric potential energy of our test charge is converted into kinetic energy as it accelerates toward the negative charge. The test charge “falls” toward the negative charge, similar to the way that an apple falls toward Earth.

Figure 5.9

Of course, there is also a repulsive force associated with electric charges. If we hold our positive test charge near another positively charged object that is stationary, there will be a repulsive force between them. The test charge still has a certain amount of electric potential energy due to its position, but when we release the test charge, it will “fall” away from the other charged object. Its electric potential energy is converted to kinetic energy as it moves away from the other charged object.

The electric potential energy that a charged object possesses due to its proximity to another charged object is proportional to the product of the charges, and inversely proportional to the distance between them.

Example 1

Calculate the electric potential energy between two oxide (O^2–) ions that are separated by a distance of 4.8 × 10^–3 m.

First, let”s find the charge on each of the oxide ions, in coulombs:

In terms of the electronics we will be studying in future chapters, we are more concerned with uniform electric fields, such as the fields between two charged plates, than we are with the electric fields surrounding point charges. Unlike the fields around point charges, which decrease over distance, uniform electric fields show the same intensity at each location. The change in the electric potential energy associated with a charged object in a uniform electric field can be calculated with the following formula:

ΔP.E._electric = –qEd

where q is the charge on the object, E is the strength of the electric field, and d is the displacement of the charged object in the direction of the field. Electric potential energy is measured in joules (J).

It is important to remember that the rules of work apply here. There is only a change in electrical potential energy if work is done on or by the electric field. In the same way that there is no change in the gravitational potential energy of a bowling ball if you simply slide it to the right side of a shelf, without changing its height, there is no change in the electrical potential energy of a charged particle if it doesn”t change its position relative to the reference point in the electric field. When the displacement is in the same direction of the electric field, we can use the formula for finding the change in electric potential energy, because the cosine of a zero degree angle is equal to 1. However, when the direction that the charged particle moves is anything other than parallel to the field, we really need to include the cosine of the angle between the field and the motion in our formula, as shown here:

ΔP.E._electric = –qEd cosθ

Example 2

Calculate the change in electrical potential energy of a proton when it moves 3.7 cm parallel to the direction of a uniform electric field with a strength of 2.5 N/C.

Figure 5.10

Note that the distance the proton travels is parallel to the electric field, and cos 0 = 1, so we don”t need to include the angle in our calculation.

You can see how the negative sign in the formula comes into play. As the proton moves in the direction of the electric field, it is analogous to an object falling toward Earth. Each of these motions results in a loss of potential energy.

Example 3

An ion of helium (He^2–) moves through a uniform electric field with a strength of 1.4 N/C. It travels 45 cm in a straight line at an angle of 127.0° from the direction of the field. Calculate the change in the electric potential energy of the ion.

Figure 5.11

Notice that our ion in Example 3 shows an increase in electric potential energy, indicated by the implied positive sign of our answer. Where did this extra energy come from? Just as a man does work on a bowling ball to raise it to a greater height, work must have been done on the helium ion to move it to a position where it possesses more electrical potential energy. The following formula summarizes the relationship between work and the change in electric potential energy.

ΔP.E._electric = –W

The negative sign for work in our formula may be easier to understand if you look at our two examples. In the first example work, was done by the field (+W), resulting in a negative change in electric potential energy. In our second example, work was done on the field (–W), and the change in electric potential energy was positive.

Combining the two formulas from this lesson, you can see that:

ΔP.E._electric = –W = –qEd cosθ

Consequently, W = qEd cosθ

Example 4

How much work would be required to move a proton (q = 1.60 × 10^–19 C) a distance 0.85 m directly against a uniform electric field with a strength of 5.6 N/C?

Note: The phrase “directly against a…field” is meant to imply that the angle between the displacement and the field is 180°.

Find: W

Solution: The work done by the field is:

W = qEd cos θ = (1.60 × 10^–19)(5.6N/C)(0.85 m)(cos 180°) = –7.6 × 10^–19 J

So, the magnitude of the required work done on the field is 7.6 × 10^–19 J.

Electric Potential

One of the factors that may contribute to the trouble that some students have with learning about electricity and magnetism is the fact that some of the terms that are used sound so similar. We just finished discussing electric potential energy, and in the next lesson we will discuss potential difference, but first, we will discuss electric potential. Electric potential sounds like electric potential energy, and although these concepts are related, they are not exactly the same thing. Electric potential is the electric potential energy that a charged object has, divided by its charge. The formula for calculating the electric potential of a point charge is:

Lesson 5–5 Review

1. _______________ is the energy that a charged object has due to its location in an electric field.

2. Calculate the electric potential energy between two protons separated by a distance of 2.8 × 10^–4 m.